arkallic
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Noob Solution Percentage Question
Hey everyone.
So this is kind of a noob question.
I'm gonna be doing a project that calls for "10ml of a 10% solution of KOH in 95% ethanol"
So here's my question. What's the right way to figure out how much KOH to put in the 10ml of ethanol?
If the ethanol is 7.89g, do I just take 10% of that (.79g)?
Or do I go by moles?
I did the math in moles and got about .94g of KOH.
So what way is right? Or with such a small difference does it even matter?
[Edited on 31-1-2014 by arkallic]
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DraconicAcid
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The 10% is almost certainly by mass (although some chemists would use mass/volume percent, which is easy, yet ridiculous), never moles. That would
give you 0.8 g of KOH, or 1 g KOH if it's mass/vol%. In most applications, it probably won't matter, because your potassium hydroxide isn't pure
anyway (commercial KOH is probably 96-98% KOH, with the rest being potassium carbonate and water)
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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arkallic
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Thanks very much!
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blogfast25
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Quote: Originally posted by DraconicAcid  | | , because your potassium hydroxide isn't pure anyway (commercial KOH is probably 96-98% KOH, with the rest being potassium carbonate and water)
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Actually, commercial KOH almost always contains 10 w% water.
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